HDU 4336 Card Collector（概率DP，状态压缩）

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1405    Accepted Submission(s): 624
Special Judge

Problem Description

In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people in the famous novel Water Margin, you will win an amazing award.

As a smart boy, you notice that to win the award, you must buy much more snacks than it seems to be. To convince your friends not to waste money any more, you should find the expected number of snacks one should buy to collect a full suit of cards.

 

Input

The first line of each test case contains one integer N (1 <= N <= 20), indicating the number of different cards you need the collect. The second line contains N numbers p1, p2, ..., pN, (p1 + p2 + ... + pN <= 1), indicating the possibility of each card to appear in a bag of snacks.

Note there is at most one card in a bag of snacks. And it is possible that there is nothing in the bag.

 

Output

Output one number for each test case, indicating the expected number of bags to buy to collect all the N different cards.

You will get accepted if the difference between your answer and the standard answer is no more that 10^-4.

 

Sample Input

1 0.1 2 0.1 0.4

 

Sample Output

10.000 10.500

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

 

 

状态压缩概率DP，或者是容斥原理、

/*
HDU 4336
题意：
有N(1<=N<=20)张卡片，每包中含有这些卡片的概率为p1,p2,````pN.
每包至多一张卡片，可能没有卡片。
求需要买多少包才能拿到所以的N张卡片，求次数的期望。


可以用容斥原理做。也可以状态压缩进行概率DP
期望DP
*/
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
const int MAXN=22;
double p[MAXN];
double dp[1<<MAXN];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        double tt=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf",&p[i]);
            tt+=p[i];
        }
        tt=1-tt;//tt就表示没有卡片的概率了
        dp[(1<<n)-1]=0;
        for(int i=(1<<n)-2;i>=0;i--)
        {
            double x=0,sum=1;
            for(int j=0;j<n;j++)
            {
                if((i&(1<<j)))x+=p[j];
                else sum+=p[j]*dp[i|(1<<j)];
            }
            dp[i]=sum/(1-tt-x);
        }
        printf("%.5lf\n",dp[0]);

    }
    return 0;
}

/*
HDU 4336
容斥原理
位元素枚举
*/
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

double p[22];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<n;i++)scanf("%lf",&p[i]);
        double ans=0;
        for(int i=1;i<(1<<n);i++)
        {
            int cnt=0;
            double sum=0;
            for(int j=0;j<n;j++)
              if(i&(1<<j))
              {
                  sum+=p[j];
                  cnt++;
              }
            if(cnt&1)ans+=1.0/sum;
            else ans-=1.0/sum;
        }
        printf("%.5lf\n",ans);
    }
    return 0;
}